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Wize High School Algebra II Textbook (Common Core) > Vectors (Extension Topic)

Vector Operations and Linear Combinations

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Vector Operations

There are two main operations we perform on vectors:
  1. Scalar multiplication
  2. Vector addition

Scalar Multiplication

For any scalar kk, scalar multiplication of a vector v⃗∈Rn\vec{v} \in \reals^n is defined as:

kv⃗ = k[v1v2⋮vn] = [kv1kv2⋮kvn]∈Rn\boxed{\quad k\vec{v} \ =\ k \begin{bmatrix} v_1\\ v_2\\ \vdots\\ v_n\\ \end{bmatrix} \ =\ \begin{bmatrix} kv_1\\ kv_2\\ \vdots\\ kv_n\\ \end{bmatrix} \quad} \in \reals^n

Geometrically, this is represented by stretching (or shrinking) the vector v⃗\vec v by a factor of kk without changing its direction.

Below we see that 2v⃗2\vec v is twice as long as v⃗\vec v while still pointing in the same direction:



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Collinear Vectors
Vectors v⃗\vec{v} and u⃗\vec{u} are collinear or parallel if there is a scalar kk such that kv⃗=u⃗k\vec{v}=\vec{u}.



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Opposite Vectors
u⃗\vec{u} and v⃗\vec{v} point in opposite directions if one vector is a negative scalar multiple of the other.
Here we see that −13u⃗-\frac{1}{3}\vec u points in the opposite direction of u⃗\vec u since we multiplied by a negative number:


Negative Vectors
−v⃗=−1v⃗-\vec{v}=-1\vec v is the negative of v⃗\vec{v}: it has the same length as v⃗\vec v, but points in the opposite direction.


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Vector Addition

For any two vectors u⃗=⟨u1,  u2,  …,  un⟩∈Rn\vec{u}=\lang u_1,\; u_2,\;\dots,\;u_n\rang \in \reals^n and v⃗=⟨v1,  v2,  …,  vn⟩∈Rn\vec{v}=\lang v_1,\; v_2,\; \dots,\; v_n\rang \in \reals^n, vector addition is defined as:

u⃗+v⃗ = [u1u2⋮un]+[v1v2⋮vn] = [u1+v1u2+v2⋮un+vn]∈Rn\boxed{\quad \vec{u}+\vec{v} \ =\ \begin{bmatrix} u_1\\ u_2\\ \vdots\\ u_n\\ \end{bmatrix} + \begin{bmatrix} v_1\\ v_2\\ \vdots\\ v_n\\ \end{bmatrix} \ =\ \begin{bmatrix} u_1+v_1\\ u_2+v_2\\ \vdots\\ u_n+v_n\\ \end{bmatrix} \quad} \in \reals^n

Watch Out!
Vectors must be in the same space to add them! E.g. A vector in R2\reals^2 cannot be added to a vector in R3\reals^3.

Geometrically, u⃗+v⃗\vec{u}+\vec{v} works by moving v⃗\vec{v} so that its tail is at the tip (arrowhead) of u⃗\vec{u}:


Vector subtraction is based on addition. It is defined as:

u⃗−v⃗=u⃗+(−v⃗)\boxed{\quad \vec{u}-\vec{v}=\vec{u}+\left(-\vec{v}\right) \quad}



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Vector Between Two Points

The vector that starts at point PP and ends at point QQ is given by:
PQ→=Q⃗−P⃗\boxed{\quad \overrightarrow{PQ}=\vec Q- \vec P \quad}

Wize Concept
Here we are using the position vectors of points PP and QQ, denoted P⃗\vec P and Q⃗\vec Q.
Recall that position vectors are used to treat points as vectors that start at the origin and end at the corresponding point.


Example
Given the points P(1,2,−3)P(1,2,-3) and Q(−2,0,4)Q(-2,0,4), find the position vector PQ→\overrightarrow{PQ}.
PQ→=Q⃗−P⃗=[−204]−[12−3]=[−2−10−24−(−3)]=[−3−27]\begin{aligned} \overrightarrow{PQ} &= \vec{Q} - \vec{P}\\ &= \begin{bmatrix} -2\\0\\4 \end{bmatrix} - \begin{bmatrix} 1\\2\\-3 \end{bmatrix}\\ &= \begin{bmatrix} -2-1\\0-2\\4-(-3) \end{bmatrix}\\ &= \begin{bmatrix} -3\\-2\\7 \end{bmatrix} \end{aligned}

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Linear Combinations

Let u⃗, v⃗∈Rn\vec{u}, \ \vec{v} \in \reals^n be vectors and let a, b∈Ra, \ b \in \reals be scalars.
A linear combination of u⃗\vec{u} and v⃗\vec{v} can be written:
au⃗+bv⃗\boxed{\quad a\vec{u}+b\vec{v}\quad}

The result is also a vector in Rn\reals^n.
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Example: Vector Operations

Given u⃗=[10−2], v⃗=[−284], w⃗=[5−2−1]∈R3\vec{u}= \begin{bmatrix} 1\\ 0\\ -2\\ \end{bmatrix} , \ \vec{v}= \begin{bmatrix} -2\\ 8\\ 4\\ \end{bmatrix} , \ \vec{w} = \begin{bmatrix} 5\\ -2\\ -1\\ \end{bmatrix} \in \reals^3 , find the resulting vector of the linear combination:
w⃗  −  2u⃗  +  12v⃗\vec{w}\;−\;2\vec{u}\;+\;\frac{1}{2}\vec{v}


w⃗ − 2u⃗ + 12v⃗ =[5−2−1]−2[10−2]+12[−284]=[5−2−1]−[2×12×02×(−2)]+[12×(−2)12×812×4]=[5−2−1]−[20−4]+[−142]=[5−2+(−1)−2−0+4−1−(−4)+2]=[225]\begin{aligned} \vec{w} \ −\ 2\vec{u} \ +\ \frac{1}{2}\vec{v} \ &= \begin{bmatrix} 5\\ -2\\ -1\\ \end{bmatrix} −\colorOne{2} \begin{bmatrix} 1\\ 0\\ -2\\ \end{bmatrix} +\colorTwo{\frac{1}{2}} \begin{bmatrix} -2\\ 8\\ 4\\ \end{bmatrix}\\ &= \begin{bmatrix} 5\\ -2\\ -1\\ \end{bmatrix} - \begin{bmatrix} \colorOne{2}\times1\\ \colorOne{2}\times0\\ \colorOne{2}\times(-2)\\ \end{bmatrix} + \begin{bmatrix} \colorTwo{\frac{1}{2}}\times(-2)\\ \colorTwo{\frac{1}{2}}\times8\\ \colorTwo{\frac{1}{2}}\times4\\ \end{bmatrix}\\ &= \begin{bmatrix} 5\\ -2\\ -1\\ \end{bmatrix} - \begin{bmatrix} 2\\ 0\\ -4\\ \end{bmatrix} + \begin{bmatrix} -1\\ 4\\ 2\\ \end{bmatrix}\\ &= \begin{bmatrix} 5-2+(-1)\\ -2-0+4\\ -1-(-4)+2\\ \end{bmatrix}\\ &= \boxed{ \begin{bmatrix} 2\\ 2\\ 5\\ \end{bmatrix}} \end{aligned}
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Example: Vector Operations

Which of the following statements are true?
  1. If u⃗, v⃗ ∈R2\vec{u},\ \vec{v}\ \in \mathbb{R}^2 , then the operation 4v⃗−12u⃗4\vec{v}-\frac{1}{2}\vec{u} is well-defined.
  2. If u⃗, v⃗ ∈R2\vec{u},\ \vec{v}\ \in \mathbb{R}^2 and w⃗ ∈R4\vec{w}\ \in \mathbb{R}^4, then the operation 2(v⃗+u⃗)−w⃗2\left(\vec{v}+\vec{u}\right)-\vec{w} is well-defined.
  3. If v⃗=−u⃗\vec{v}=-\vec{u}, then u⃗\vec{u} and v⃗\vec{v} are collinear.
  4. u⃗=⟨−1, 2, −3, −4⟩\vec{u}=\lang-1,\ 2,\ -3,\ -4\rang and v⃗=⟨1, −2, 3, −4⟩\vec{v}=\lang1,\ -2,\ 3,\ -4\rang are parallel and opposite vectors.
1. TRUE: Scalar multiplication is always valid. Since the two vectors are in the same space, R2\mathbb{R}^2, vector addition/subtraction is also well-defined.

2. FALSE: Vector addition/subtraction between vectors in different spaces is never valid.

3. TRUE: If one vector is the negative of another vector, that means they are a scalar multiples of one another (factor k=−1k=-1), hence they are collinear.

4. FALSE: The first three components are indeed scalar multiples of one another (by a factor of -1); however, the last component fails to follow this rule. The vectors are therefore not scalar multiples of one another, thus they are neither parallel nor opposite vectors.

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Example: Vector Operations

P(1,1)P(1,1), Q(−3,5)Q(-3,5), and R(x,y)R(x,y) are points in R2\mathbb{R}^2, and QR→=−12QP→\overrightarrow{QR}=-\frac{1}{2}\overrightarrow{QP}.
Find the coordinates of the point R(x,y)R(x,y).

Since we know both PP and QQ:
QP→=[1−(−3)1−5]=[4−4]\overrightarrow{QP}= \begin{bmatrix} 1-(-3)\\ 1-5\\ \end{bmatrix}= \begin{bmatrix} 4\\ -4\\ \end{bmatrix}

Now we can use the relationship between QR→\overrightarrow{QR} and QP→\overrightarrow{QP}:
QR→=−12QP→=−12[4−4]=[−22]\overrightarrow{QR} =-\frac{1}{2}\overrightarrow{QP} =-\frac{1}{2} \begin{bmatrix} 4\\ -4\\ \end{bmatrix} =\boxed{\begin{bmatrix} -2\\ 2\\ \end{bmatrix}}

And by the formula for a vector between two points, we can also write:

QR→=[x−(−3)y−5]=[x+3y−5]\overrightarrow{QR} =\begin{bmatrix} x-(-3)\\ y-5\\ \end{bmatrix} =\boxed{ \begin{bmatrix} x+3\\ y-5\\ \end{bmatrix}}

The boxed expressions are both representations of QR→\overrightarrow{QR} so they must be equal:
[x+3y−5]=[−22]\begin{bmatrix} \colorOne{x+3}\\ \colorTwo{y-5}\\ \end{bmatrix} = \begin{bmatrix} \colorOne{-2}\\ \colorTwo{2}\\ \end{bmatrix}

This gives us two equations (one for each component):

x+3=−2  ⟹  x=−5\colorOne{x+3=-2} \quad\implies\quad x=-5

y−5=2     ⟹  y=7\colorTwo{y-5=2} \ \ \ \quad\implies\quad y=7

∴  R(−5,7)\therefore\ \ \boxed{R(-5,7)}

Practice: Collinear Vectors

Find the value of cc such that the vectors u⃗=⟨−3,c,3⟩\vec u = \lang −3,c,3 \rang and v⃗=⟨1,−2,−1⟩\vec v = \lang 1,−2,−1 \rang are collinear.

Practice: Collinear Vectors

Given u⃗=⟨2,−5,6⟩,  v⃗=⟨−4,10,−12⟩,\vec{u}=\lang 2,−5,6\rang ,\ \ \vec{v}=\lang−4,10,−12\rang , and 𝑤⃗=⟨−2,5,−6⟩𝑤⃗=\lang −2,5,−6 \rang, select all statements that are true.

Practice: Vector Operations

A parallelogram has sides ABAB, BCBC, CDCD and DADA.

The following coordinates are known: A(2,0,1)A(2,0,1), C(6,0,0)C(6,0,0), and M(3,1,0)M(3,1,0), where MM is the midpoint between AA and BB.

Find the vector BD→\overrightarrow{BD}.